What property of the electron causes electron domains to have an effect on molecular shapes?

Short Answer

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The property of electrons that causes electron domains to have an effect on molecular shapes is their negative charge. This leads to repulsion between electron domains, governed by the Valence Shell Electron Pair Repulsion (VSEPR) theory, which in turn dictates the overall geometry of molecules to minimize these repulsive forces.

Step by step solution

01

Understand electron domains

Electron domains are regions around an atom where valence electrons are most likely to be found. They include both bonding and non-bonding electron pairs. The electron pairs occupy space around the central atom, and they repel each other in an attempt to minimize the repulsive forces between them. This repulsion between electron domains determines the overall geometry of the molecule.
02

Electron properties responsible for domain repulsion

The property of electron responsible for the repulsion between electron domains is their negative charge. Electrons carry a negative charge, and as a result, they repel other negatively charged particles, such as other electrons. This repulsion is governed by the principle known as the Valence Shell Electron Pair Repulsion (VSEPR) theory, which states that electron pairs in the valence shell of an atom repel one another, and they arrange themselves in a way to minimize these repulsive forces.
03

How electron repulsion affects molecular shape

The repulsion between electron domains plays a significant role in determining the molecular shape. In an attempt to minimize the repulsive forces, the electron domains occupy positions that maximize the angles between them and dictate the molecular geometry. For example, if a central atom has two electron domains around it, they will arrange themselves linearly with an angle of 180 degrees. If there are three electron domains, they will arrange in a trigonal planar shape with angles of 120 degrees, and so on. In conclusion, the property of electrons that causes electron domains to have an effect on molecular shapes is their negative charge. This leads to repulsion between electron domains, which in turn dictates the overall geometry of molecules to minimize these repulsive forces.

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Most popular questions from this chapter

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}\) is a planar molecule that has the following carbocarbon bond lengths: (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(C-\) bond length in butadiene \((1.48\) A) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length \((1.54 \hat{\mathrm{A}}) .\) Does this imply that the middle \(\mathrm{C}-\mathrm{Cbond}\) in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\)? (\mathbf{d} ) Based on your answer for part ( c ),discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\) C bond.

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$\mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6}\), you run into a problem. What is it? (c) What could you do to resolve the difficulty in part \((\mathrm{b}) ?(\mathbf{d})\) The molecule \(\mathrm{IF}_{7}\) has a pentagonalbipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\).

(a) Starting with the orbital diagram of a boron atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{BF}_{3}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Are any valence atomic orbitals of \(\mathrm{B}\) left unhybridized? If so, how are they oriented relative to the hybrid orbitals?

(a) Write a single Lewis structure for \(\mathrm{SO}_{3}\), and determine the hybridization at the \(\mathrm{S}\) atom. (b) Are there other equivalent Lewis structures for the molecule? (c) Would you expect \(\mathrm{SO}_{3}\) to exhibit delocalized \(\pi\) bonding? Explain.

In ozone, \(\mathrm{O}_{3},\) the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

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